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Math Help - Trig Limit Help

  1. #1
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    Trig Limit Help

    1. limx->0 (cos(x)-1)/sin(2x)

    I have it down to (-sinx^2)/((sin2x)(1+cosx)), but I can't seem to finish it

    2. For my second question I don't even have a start. Here it is.

    limx->0 (ln(1+x)/(sin3x))

    Thank you so much in advance
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  2. #2
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    Quote Originally Posted by chrisski View Post
    1. limx->0 (cos(x)-1)/sin(2x)

    I have it down to (-sinx^2)/((sin2x)(1+cosx)), but I can't seem to finish it
    I'm not sure how you arrived at this but maybe this will help.

    \frac{-\sin^2(x)}{(\sin(2x))(1+\cos(x))}

    \frac{-\sin^2(x)}{(2\sin(x)\cos(x))(1+\cos(x))}

    \frac{-\sin(x)}{(2\cos(x))(1+\cos(x))}
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  3. #3
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    I could very well have arrived at an incorrect point. If I'm doing it incorrectly please inform me. Also, any help on the second one would be appreciated.
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  4. #4
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    Quote Originally Posted by chrisski View Post
    I could very well have arrived at an incorrect point. If I'm doing it incorrectly please inform me. Also, any help on the second one would be appreciated.
     \lim_{x \to 0} \frac{ln(1+x)}{\sin 3x}=\lim_{x \to 0} \left ( \frac{ln(1+x)}{\sin 3x} \cdot \frac{3x}{3x} \right )
     = \lim_{x \to 0}  \left  ( \frac{ln(1+x)}{ 3x} \cdot \frac{3x}{\sin 3x} \right )
    =\frac{1}{3} \lim_{x \to 0}  \left (\frac{ln(1+x)}{ x} \right )  \cdot \lim_{x \to 0}  \left (\frac{1}{\frac{\sin 3x}{3x}} \right )
    =\frac{1}{3} \cdot 1 \cdot 1 = \frac{1}{3}
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  5. #5
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    Quote Originally Posted by pickslides View Post
    I'm not sure how you arrived at this but maybe this will help.

    \frac{-\sin^2(x)}{(\sin(2x))(1+\cos(x))}

    \frac{-\sin^2(x)}{(2\sin(x)\cos(x))(1+\cos(x))}

    \frac{-\sin(x)}{(2\cos(x))(1+\cos(x))}
    \frac{cosx-1}{sin2x}=\frac{cosx-1}{sin2x}\cdot \frac{cosx+1}{cosx+1} = \frac{(cosx-1)(cosx+1)}{sin2x(cosx+1)} =  \frac{cos^2(x)-1}{sin2x(cosx+1)} = \frac{-sin^2(x)}{sin2x(cosx+1)}
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